Question

Simplify $\frac{sin^2x+cos^2x}{sin x}$

Answer 1

Let's see

$\\ \rm\Rrightarrow \dfrac{sin^2x+cos^x}{sinx}$

• sin²x+cos²x=1

$\\ \rm\Rrightarrow \dfrac{1}{sinx}$

$\\ \rm\Rrightarrow cosecx$

Answer 2

Answer:

cosec x

(or csc x)

Step-by-step explanation:

Trig Identities

$\sf \sin^2 x + \cos^2 x=1$

$\sf \dfrac{1}{\sin x}=cosec\: x$

Therefore,

$\begin{aligned}\implies \sf \dfrac{\sin^2x + \cos^2x}{\sin x} &= \sf\dfrac{1}{\sin x}\\\\ & = \sf cosec\:x \end{aligned}$